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Stability for one-mode solution of Maxwell MHD equation with chiral anomaly
Let's discuss stability of exact one-mode solution of Maxwell equations with presense of chiral anomaly and in hydrodynamical approximation. Corresponding system of equations has the form \ \times \mathbf E = -\partial_{0}\mathbf B \qquad (1) , \ \times \mathbf B = \sigma \mathbf E + \frac{\alpha}{\pi}\mu_{5}\mathbf B \qquad (2) , \ \partial_{0}\mu_{5} = \frac{\alpha }{\pi f_{5}^{2}}(\mathbf E \cdot \mathbf B) \qquad (3) . For one-mode solution, \ \mathbf B(t) = b(t)\mathbf b_{k}, \quad \mathbf b_{k} = (sin(kz), cos(kz), 0), \quad b(0) = B_{0} . Then we have from \ (1)-(3) \ \mathbf E = -\frac{1}{k}\dot{b}(t)\mathbf b_{k}, \quad \partial_{0}\mu_{5} = -\frac{\alpha }{2\pi f_{5}^{2}k}\dot{b}^{2}(t) \Rightarrow \mu_{5} = \mu_{5}(0) + \frac{\alpha B_{0}^{2}}{2 \pi kf_{5}^{2}} - \frac{\alpha b^{2}(t)}{2 \pi kf_{5}^{2}} , \ kb(t) = -\frac{\sigma}{k}\dot{b}(t) + \frac{\alpha}{\pi}\left( \mu_{5}(0) + \frac{\alpha B_{0}^{2}}{2 \pi kf_{5}^{2}} - \frac{\alpha b^{2}(t)}{2 \pi kf_{5}^{2}}\right)b(t) . For \ \frac{\alpha}{\pi}\mu_{5}(0) \neq k we have solution \ b(t) = \frac{B_{0}\sqrt{\gamma B_{0}^{2} + \delta }e^{(B_{0}^{2}\gamma + \delta ) t}}{\sqrt{ \gamma B_{0}^{2}e^{2(B_{0}^{2}\gamma + \delta ) t} + \delta}} , \quad \delta = \frac{\alpha \mu_{5}^{0}}{\pi} - k , \quad \gamma = \frac{\alpha^{2}}{2\pi^{2}\sigma f_{5}^{2}}, \quad \delta + B_{0}^{2}\gamma > 0 \qquad (4) . From \ (4) we see that \ b(\infty ) \to B_{0}\sqrt{1 + \frac{\delta }{\gamma B_{0}^{2}}} , \quad \mathbf E (\infty ) \to 0, \quad \mu_{5} (\infty ) \to \frac{\pi k }{\alpha} . Perturbations Let's perturb solution of \ (1)-(3) and then linearize system on perturbations: the result is \ \partial_{0}\delta \mu_{5} = \gamma \left( (\delta \mathbf E \cdot \mathbf B ) + (\delta \mathbf B \cdot \mathbf E )\right) \qquad (5) , \ \times \delta \mathbf E = -\partial_{0}\delta \mathbf B \qquad (6) , \ \times \delta \mathbf B = \sigma \delta \mathbf E + \frac{\alpha}{\pi}\left( \mu_{5}\delta \mathbf B + \mathbf B \delta \mu_{5}\right) \qquad (7) . Let's take curl and divergence of \ (6)-(7) : \ (\nabla \cdot \delta \mathbf E ) = -\frac{\alpha}{\pi \sigma}(\mathbf B \cdot \nabla \delta \mu_{5}) \qquad (8) , \ \nabla (\nabla \cdot \mathbf E) - \Delta \delta \mathbf E = - \partial_{0}\times \delta \mathbf B \qquad (9) \ -\Delta \delta \mathbf B = -\sigma \partial_{0}\delta \mathbf B + \frac{\alpha}{\pi}\left( \mu \times \delta \mathbf B + \delta \mu_{5} \times \mathbf B + \delta \mu_{5}\times \mathbf B\right) \qquad (10) . From these equations we have \ \delta \mathbf E = -\frac{\alpha }{\pi \sigma }\delta \mu_{5}\mathbf B + \delta \tilde {\mathbf E}, \quad \nabla \cdot \delta \tilde{\mathbf E} = 0 , \ \times \delta \mathbf B = \frac{\alpha}{\pi}\mu \delta \mathbf B \Rightarrow \Delta \delta \mathbf B = -\left( \frac{\alpha \mu}{\pi \sigma}\right)^{2}\delta \mathbf B \Rightarrow \delta \mathbf B = \mathbf A cos (\frac{\alpha \mu}{\pi }(\mathbf n \cdot \mathbf r)) , where \ (\mathbf n \cdot \mathbf A ) = 0, \quad \mathbf A = const, \quad \mathbf n^{2} = 1 . This is oscillating solution, so it doesn't grow. We have derived that trivial solution for magnetic field \ \delta \mathbf B is stable (not asymptotically). Let's now derive the expressions for \ \delta \mathbf E , \delta \mu_{5} . From \ (6) we have \ \times \delta \mathbf E = \frac{\alpha }{\pi}\partial_{0}\mu_{5}\mathbf A sin\left(\frac{\alpha \mu_{5}}{\pi} (\mathbf n \cdot \mathbf r)\right)(\mathbf n \cdot \mathbf r) . We may set, for simplicity, \ \mathbf A = (1, 0, 0), \mathbf n = (0, 0, 1) . Then \ \delta \mathbf E = \delta \mathbf E(z) , and \ \delta \mathbf E = \frac{\alpha }{\pi}\partial_{0}\mu_{5} \left(0, -\frac{sin\left(\frac{\alpha \mu_{5}}{\pi} z\right)}{\left(\frac{\alpha \mu_{5}}{\pi}\right)^{2}} + \frac{cos\left(\frac{\alpha \mu_{5}}{\pi} z\right)}{\frac{\alpha \mu_{5}}{\pi}}, 0\right) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \